B. A Continuous Model of Newton's Law of Cooling
Introduction
Ever since humans began to build shelters the problem of escaping heat has been a chief concern. Engineers and scientists are continually working on new ideas for manufacturing low cost, space efficient materials for thermal insulation. Thermal insulation has a variety of uses from clothing to the tiles on the space shuttle, and one of its biggest uses is in the homeconstruction industry. Today virtually all new housing incorporates thermal insulation in attics and walls to reduce heating and cooling costs. Factors affecting heat loss include radiation (loss of heat in the air around an object), convection (loss of heat due to the movement of air on either side of an object), and conduction ( loss of heat due to contact between particles making up an object). In this module students investigate the heat flow problem from two perspectives: 1) the rate of flow of heat involved in cooling or heating an object or a liquid and 2) the effort to keep a home insulated at minimum cost.
Newton's Law of Cooling and Heating
We are all aware of instances in which a coroner is required to determine the approximate time of death of a homicide victim. Knowing something about how fast the temperature of a human body cools down from 98.6oF to room temperature, about 70oF, can be of significant aid in the coroner's conclusion. A law of physics useful in such cases is called Newton's Law of Cooling.
Suppose you heat a pie in an oven to 450oF, then remove it and let it cool in a kitchen in which the air temperature is 70oF. If you construct a graph of the temperature of the pie vs. the time in minutes, what kind of curve do you get? During what time intervals does the pie cool most quickly? During which time intervals does it cool least quickly? What is the low limit for the pie's temperature? Try your hand at sketching such a graph. What is an "asymptote" for the curve you drew? Newton's Law of Cooling states that the change in temperature of the pie from minute to minute is proportional to the difference between the temperature of the pie and the temperature of the room. In symbols:
Tn+1 - Tn = k(Tn - 70) k < 0.
Note that the constant of proportionality k is negative since the temperature is decreasing. In general, if the room temperature is any constant R, then the equation for Newton's Law becomes:
(1) Tn+1 - Tn = k(Tn - R) k < 0
or
Tn+1 = Tn + k(Tn - R) k < 0.
Testing Newton's Law of Cooling
In the following experiment you are to determine whether Newton's Law of Cooling holds for a cup of hot coffee in an insulated cup with a lid.
1. Collect temperature data at two-minute intervals for at least 50 minutes.
2. Plot a graph of the temperature data over time. Describe the curve. Does your
curve suggest an asymptote?
3. Test to see if equation (1) holds by graphing (Tn+1 - Tn) vs. (Tn - R). If Newton's
Law holds for your data, what kind of curve should you get? How can you
approximate the constant of proportionality k? (Hint: find the slope of a regression
line.)
Notice that equation (1) above is an example of the difference equation involved in the exponential growth model discussed in sections 2.3 - 2.5 in the Berry and Houston text. Using the general solution of that model given in section 2.4, page 43, show that the general solution to the difference equation (1) is :
(2) Tn = R + (T0 - R)(1 + k)n
where T0 represents the original temperature of the liquid. Equation (2) indicates theoretically how the temperature of the coffee should change from minute to minute under Newton's Law of Cooling. Now perform error analysis to determine how closely your data approximates that generated by the theoretical equation.
A. Tutorial Problems
1. Suppose a pie is baking in a 350oF oven. The pie is removed when its temperature is 180oF and is left to cool in a kitchen at 70oF. After 10 minutes , the temperature of the pie is 125oF. Find the temperature of the pie after 15 minutes. How long does it take the pie to cool down to 75oF?
2. Mr. Smith's body was found is his kitchen at 9:00 A.M. at whidch time his body temperature was 77.3oF and the room temperature was 70oF. One hour later the coroner arrived and found the body temperature was was 76.1oF. Assuming that Mr. smith's body temperature was the normal 98.6oF at the time of his death, at what time was he murdered?
3. Newton's Law of Heating is a corresponding principle which applies if an object is being warmed rather than cooled. The same formulas apply except the constant of proportionality is positive in the warming case. Use Newton's Law of Heating to solve the following problem: A chicken is removed from the refrigerator at a temperature of 40oF and placed in an oven kept at the constant temperaturre of 350oF. After 10 minutes the temperature of the chicken is 70oF. The chicken is considered cooked when its temperature reaches 180oF. How long must it remain in the oven?
B. A Continuous Model of Newton's Law of Cooling
As in Tutorial Problem #13, Section 2.6, P. 38, Berry & Houston, Newton's Cooling Law model can be formulated in terms of differential equations. Show that the difference equation model in equation (1) becomes:
(3) dT = k(T - R) k < 0
dt
where T(t) represents the temperature continuously at any time t, and that the solution to the differential equation is:
(4) T = R + (T0 - R)ekt .
Now return to your original coffee data and test to see if your data conforms to the continuous exponential model in equation (4). Procedure: Linearize the relationship in equation (4) by taking logarithms and use linear regression to find the best values of the parameters in the "least squares" sense.
C. Insulating a House
It is impossible to build an airtight house, one which retains all of its heat. A house loses heat through doors, windows, floors, etc. It is estimated that a house without insulation loses 25% of its heat through the roof, 30% through its walls, 10% through its windows, and the remaining 35% through the floor, gaps around the windows and doors, etc. In this section we consider two ways to reduce heat loss in a house: 1) Installing double window panes referred to as "double glazing" and 2) filling the cavities between the walls with insulation which we refer to as "cavity-wall insulation." Our problem is to compare the relative cost effectiveness of the one method against the other. Read Case Study 6, P. 84, Berry & Houston for additional details and suggestions relative to this problem.
Creating a Model
A basic model for the transfer or loss of heat per unit time (measured in watts) through a window or door is:
(5) H = UxA(TI - T0)
where A is the area of the window or door, TI is the temperature inside and T0 is the temperature outside of the window or door, and U is a constant which is known for different materials. (See Table I below for typical U-values.) To compare the cost effectiveness of one method versus the other we use the following equation to determine the heat saved from either method:
(6) Heat saved by insulation = (heat loss when noninsulated) - (heat loss when insulated).
Next we approach the problem of comparing money saved by one method versus the other. We define the following variables (See Case Study 6, P. 93.):
AG = area of glass AB = area of walls
HG = heat saved by glazing HB = heat saved by wall insulation
CG = cost of glazing CB = cost of wall insulation
SG = money saved per unit time glazing SB = money saved per unit time wall insul.
c = cost of heat per unit time
From (5) and (6) we can write
For Windows:
(7) HG = UNAG (TI - T0) - UIAG (TI - T0) SG = cHG
where UN = the U value for single pane (noninsulated) windows and
UI = the U value for double glazed (insulated) windows.
For Walls:
(8) HB = U1NAB (TI - T0) - U1IAB (TI - T0) SB = cHB
where U1N = the U value for unfilled (noninsulated) cavity walls and
U1I = the U value for filled (insulated) cavity walls.
A Strategy for Cost Effectiveness
To facilitate a cost/savings comparison between double glazing and cavity-wall insulation, we define a quantity P which we call "payback period" for each type of insulation.
P = the ratio of the cost of the insulation to the money saved by installing the insulation. Thus,
for windows
CG
PG
SG
and for walls
CB
PB
SB
where CG = the cost of double glazing and CB = the cost of wall-cavity filling. It is clear from the definition that one desires as small a payback period as possible. Thus a model for cost effectiveness is formulated as follows:
If PB < PG then one should install cavity wall insulation
If PG < PB then one should install double glazing
One final computation permits us to relate payback period to cost alone. Using the definitions of payback period, equations (7) and (8), and Table I below, we have:
PG CG SB CG c (U1N - U1I)AB (TI -T0) CG AB 0.0726
PB SG CB CB c (UN - UI)AG (TI - T0) CB AG
Now if cG represents the cost of double glazing per unit area, and cB represents the cost of cavity wall insulation per unit area then, cG = CGAG and cB = CBAB from which we get
PG cG
0.0726
PB cB
This equation indicates that the cost effectiveness test depends only on the cost per unit area of the two types of insulation.
Tutorial Problems
1. Use Table II below to determine which method is more cost effective: double glazing or cavity wall insulation.
2. Which of the double glazing systems listed in Table II becomes cost effective if we adopt the strategy of P < 2P? P < 3P?
TABLE I
Typical U-values for Walls and Windows
Material |
U-value |
Brick Wall (no cavity) |
1.92 |
Brick wall (with cavity not filled) |
0.873 |
Brick Wall (with filled cavity) |
0.5 |
Single Glass Pane (6 mm) |
6.41 |
Double Glazed Window |
1.27 |
TABLE II
Typical Insulation Cost
Type of Insulation |
Cost Per Square Meter (Pounds) |
Professionally fitted sealed double glazed replacement windows |
181 |
Professionally fitted secondary units |
28.9 |
Do-it-yourself secondary double glazing |
16 |
Do-it-yourself sealed double glazed replacement windows |
30 |
Cavity wall insulation |
2 |
D. Insulating a Roof
The major loss of heat in many houses is through the roof. The loss of heat can be reduced by installing a layer of insulation above the ceiling. With no insulation, the cost of heating could be high. With a large amount of insulation the heat loss is reduced, but the insulation cost might be too high. The excess money might have been invested in an interest bearing account. The loss of that interest could be considered a "cost." The problem in this section is to find the optimum thickness of roof insulation which minimizes total heating cost.
Cost Analysis
In a steady state situation, total costs are given by:
Total Cost = Cost of Heating + Interest Loss Per Year.
(a) (b)
We use the following parameters:
B($) = Cost of 1 Kw/hr
r(%) = Interest rate per annum
C($) = Cost of insulation per cubic meter
A(m2) = Area of ceiling in square meters
x(m) = Thickness of the insulation
H(hr) = Number of heating hours in 1 year
Q(Kw/hr) = Heat lost through the roof surface
(a) Cost of heating per year = Heat lost x Cost per Kw/hr x Heating Period for 1 year
= QBH
(b) The interest lost per year = CxAr/100.
Combining (a) and (b) yields an equation for total cost denoted by F:
(1) Total Cost F = QBH + CxAr/100
Heat Loss Analysis
Fourier's Law gives the heat lost through an area of surface A as:
or
where k is the thermal conductivity coefficient and T(x) is the temperature x meters from the roof. The general solution to this differential equation is:
where c is an arbitrary constant. If we assume that the room temperature is 25oC and that the average outside temperature is 10oC, we can use these two boundary conditions to determine that c=25 and show that the heat lost through surface area A is:
Now if values of the parameters are given, we can use (1) to write down an equation for total cost F in terms of x, the thickness of the insulation.
Tutorial Problems:
1. Assume that in a certain house, the heat is on for 200 days for an average of 12 hours per day and that the roof is 700 square meters in surface area. The cost of 1Kw/hr is $0.005 and the cost of insulation is $500 per cubic meter, and the thermal conductivity for the insulation k=2x10-5. Finally, assume that interest can be earned on money invested at 8% per annum. Use (1) above to write an equation F(x) which gives total cost of insulating the roof in terms of x, the thickness of the insulation.
2. Sketch a graph of F(x) and estimate from the graph the thickness x which will minimize the total cost.
3. Use calculus to find the optimum value of x analytically and compare with your graphical estimate. (Hint: take the first derivative of F(x)).